104 - Arbitrage

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ubernewb
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104 WA

Post by ubernewb » Mon May 01, 2006 8:39 pm

Hey,
I'm trying to solve the arbitrage problem, but it's not behaving as expected. It's worked fine for all the sample data I found on this forum, but I'm still getting a WA on submit. Can someone help me? It's C++, by the way.

Code: Select all

// Arbitrage

#include <iostream>
#include <sstream>
#include <stack>
using namespace std;

#define MAX 20
#define EPS 0.000001

string intToString(int x)
{
	ostringstream ss;
	ss << x;
	return ss.str();
}

// Floyd-Warshall Algorithm
string FW(double table[MAX][MAX], int numCountries)
{
	string output;

	// Initialize the tables
	double best[MAX][MAX][MAX];
	int path[MAX][MAX][MAX];
	for (int i = 0; i < numCountries; i++)
		for (int j = 0; j < numCountries; j++)
			for (int k = 0; k < numCountries; k++)
			{
				best[i][j][k] = 0;
				if (k == 1)
				{
					best[i][j][k] = table[i][j];
					path[i][j][k] = i;
				}
			}

	// Begin the search
	for (int steps = 2; steps <= numCountries; steps++)
	{
		for (int k = 0; k < numCountries; k++)
		{
			for (int i = 0; i < numCountries; i++)
				for (int j = 0; j < numCountries; j++)
				{
					double temp = best[i][k][steps-1] * best[k][j][1];
					if (temp > best[i][j][steps])
					{
						best[i][j][steps] = temp;
						path[i][j][steps] = k;
					}
				}
		}
		for (int i = 0; i < numCountries; i++)
		{
			if (best[i][i][steps] - 1.01 > EPS)
			{
				stack<int> s;
				s.push(i);
				int next = i;
				for (int j = steps; j > 0; j--)
				{
					next = path[i][next][j];
					s.push(next);
				}
				output = intToString(s.top() + 1);
				s.pop();
				while(!s.empty())
				{
					output += " ";
					output += intToString(s.top() + 1);
					s.pop();
				}
				output += "\n";
				return output;
			}
		}
	}
	return "no arbitrage sequence exists\n";
}

string findSeq(double table[MAX][MAX], int numCountries)
{
	double best[MAX][MAX];
	return FW(table, numCountries);
}

int main()
{
	string input, output;
	int numCountries;
	while (true)
	{
		getline(cin, input);
		if (input == "")
		{
			break;
		}
		numCountries = atoi(input.c_str());
		double table[MAX][MAX];
		for (int row = 0; row < numCountries; row++)
		{
			getline(cin, input);
			istringstream *ss = new istringstream;
			ss->str(input);
			for (int col = 0; col < numCountries; col++)
			{
				if (row == col)
				{
					table[row][col] = 1.0;
				}
				else
				{
					*ss >> table[row][col];
				}
			}
			delete ss;
		}
		output += findSeq(table, numCountries);
	}
	cout << output;
}

snublefot
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Post by snublefot » Wed May 03, 2006 10:30 am

It worked fine with my solution.

snar
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Location: Yerevan, Armenia

Post by snar » Wed May 03, 2006 7:25 pm

Hi, here's my output for your input.

10 16 10 16 10
1 2 1
1 2 4 1
no arbitrage sequence exists
5 6 5
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 1
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 1
no arbitrage sequence exists
no arbitrage sequence exists
Narek Saribekyan

Dai Ming
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Location: China

Post by Dai Ming » Sun May 14, 2006 3:21 am

10 16 10 16 10
1 2 1
1 2 3 1
no arbitrage sequence exists
5 6 5
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 1
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 1
no arbitrage sequence exists
no arbitrage sequence exists

this is the output of my program ,is it right?
especially the third one 1 2 3 1
It can't got AC But I do not know the prblom.
Help~~~~~~~~~~~~~~~~

chunyi81
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Joined: Sat Jun 21, 2003 4:19 am
Location: Singapore

Post by chunyi81 » Sun May 14, 2006 3:27 pm

My AC code gave the following output for the test case in this thread:

10 16 10 16 10
1 2 1
1 4 3 1
no arbitrage sequence exists
5 6 5
no arbitrage sequence exists
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 1
no arbitrage sequence exists
1 20 1 20 1 20 1 20 1 20 1 20 1 20 1 20 1 20 1 20 1

This problem has a special corrector judge, which means there can be more than one arbitrage sequence for a given test case. The output from my program for the third case is different as well.

Dai Ming
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Joined: Sun Apr 23, 2006 3:54 am
Location: China

Post by Dai Ming » Mon May 15, 2006 6:41 am

10 16 10 16 10
1 2 1
1 2 3 1
no arbitrage sequence exists
5 6 5
no arbitrage sequence exists
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 1
no arbitrage sequence exists
1 20 1 20 1 20 1 20 1 20 1 20 1 20 1 20 1 20 1 20 1

this is my new result of my program,Is there any problem with it.
I think it is all right,but It can't get AC.
I am so puzzled.

Code: Select all

#include <iostream>
#include <stdio.h>
#include <math.h>
#include <fstream>

using namespace std;

double r,rate[20][20];
bool flag,tf;
int s[21],sf[21];
int time,t,d;

void next()
{
	int i,j,k;	

		for( i = time - 1; i >= 0; i --)
		{
			if(s[i] < d - (time - i))
			{
				s[i]++;
				for(j = i + 1; j < time; j ++)
				{
					s[j] = s[j - 1] + 1;
				}
				s[time] = s[0];
				return;
			} else
			{
				if(i > 0)
				{
					continue;
				} else
				{
					if(time < d)
					{
						time++;
						if(time >= t)
						{
							tf = true;
							
							return;
						}

						for(k = 0; k < time; k++)
						{
							s[k] = k;
						}
						s[time] = 0;
					} else
					{
						flag = true;
						return;
					}
				}
			}
		}		
}

void fang()
{	
	int i,j,k;	
	i = log10(1.01)/log10(r);
	
		 i++;		
		if(i*time <=d)
		{			
			for(j= 0; j < i; j++)
			{
				for(k = 0; k < time; k++)
					sf[j*time + k] = s[k];
			}
			t = i*time;
			sf[t] = s[0];
			tf = true;			
		}	
}

int main()
{
	int i,j;
	bool h; 
	int f = scanf("%d",&d);
	while(1 == f)
	{
		flag = false;
		tf = false;
		t = 21;
		for(i = 0; i < d; i ++)
		{
			for(j = 0; j < d; j ++)
			{
				if(i == j)
				{
					rate[i][j] = 1.0;
				}else
				{
					scanf("%lf",&rate[i][j]);
				}
			}
		}
		///////////////////////////////////////		
		time = 2;
		s[0] = 0;
		s[1] = 1;
		s[2] = 0;		
		
		h = true;
		while(1)
		{		
			r = 1.0;
			h = true;
			for( i = 0; i < time; i ++)
			{
				r *= rate[s[i]][s[i+1]];
			}			 
			
			if(tf) break;
			if( r >= 1.01)
			{
				cout<<" r - 1.01"<<endl;
				h = false;
				break;
			}
			else if( r > 1.0 && time < 11 )
			{				
				fang();
			}
			else if(flag)
			{
				break;
			}
			next();			
		}
		if(tf)
		{
			for(i = 0; i < t; i++)
			{
				cout<<sf[i]+1<<" ";			
			}
			cout<<sf[i]+1<<endl;					
		}
		else if(h && flag)
		{
			cout<<"no arbitrage sequence exists"<<endl;	
					}
		else
		{
			for(i = 0; i < time; i++)
			{
				cout<<s[i]+1<<" ";	
				
			}
			cout<<s[time]+1<<endl;	
					
		}

		f = scanf("%d",&d);
	}
	return 0;
}



LithiumDex
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Post by LithiumDex » Mon Jun 26, 2006 2:41 am

Here is my output for this data set:

10 16 10 16 10
1 2 1
no arbitrage sequence exists ** (ERROR)
no arbitrage sequence exists
5 6 5
no arbitrage sequence exists
no arbitrage sequence exists ** (ERROR)
no arbitrage sequence exists
1 20 1 20 1 20 1 20 1 20 1 20 1 20 1 20 1 20 1 20 1

And here is my code:

Code: Select all

#include <iostream>
#include <vector>
#include <algorithm>
#include <string>
#include <sstream>
#include <math.h>
using namespace std;

#define T(_I, _J) table[_I+(_J<<5)]
#define C(_I, _J) cost[_I+(_J<<5)]
#define N(_I, _J) next[_I+(_J<<5)]

double weight ( double v )
{
    return 1.0-(log(v)/10000.0);
}

int main ( void )
{
    int n;
    double table[32<<5];
    double cost[32<<5];
    int next[32<<5];
    
    while (cin >> n)
    {
        int i, j, k;
                
        for (i=0; i<n; i++)
            for (j=0; j<n; j++)
            {
                N(i, j) = j;
                
                if (j==i)
                {
                    T(i, j) = 1.0;
                    C(i, j) = 100000;
                }
                else
                {
                    double in;
                    cin >> in;
                    T(i, j) = in;
                    C(i, j) = weight(in);
                }
            }
            
        for (k=0; k<n; k++)
            for (i=0; i<n; i++)
                for (j=0; j<n; j++)
                {
                    double sum = C(i, k) + C(k, j);
                    
                    if (sum < C(i, j))
                    {
                        C(i, j) = sum;
                        N(i, j) = N(i, k);
                    }
                }
                
        vector<int> paths[20];
        int minsteps = n+1;
        int mini = -1;
                
        for (i=0; i<n; i++)
        {
            int w = i;
            int l;
            double a = 1.0;
            int steps = 0;
            
            while (true)
            {
                paths[i].push_back(w);
                steps++;
                
                if (steps > n)
                    break;
                
                l = w;
                w = N(l, i);
                a *= T(l, w);
                
                if (w == i)
                {
                    if (a <= 1.0)
                        break;
                    else if (a > 1.01)
                    {
                        paths[i].push_back(w);
                        
                        if (steps < minsteps)
                        {
                            minsteps = steps;
                            mini = i;
                        }
                        
                        break;
                    }
                }
            }
        }
        
        if (mini == -1)
        {
            cout << "no arbitrage sequence exists" << endl;
        }
        else
        {
            for (i=0; i<(paths[mini].size()-1); i++)
                cout << paths[mini][i]+1 << " ";
            cout << paths[mini][paths[mini].size()-1]+1 << endl;
        }
        
    }
    
    return 1;
}
I know my problem is with how I'm weighting each number, (and maybe that I am weighting each number), however, I tried two different methods of doing this, the method you see there, which produces those results, and a counting the digits which fixed (atleast #3) of the outputs, and broke (atleast #1) of the outputs, I tried combining the two methods, but with no success, any ideas?
- Chris Adams

C
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Thanks

Post by C » Mon Jun 26, 2006 7:29 pm

This hint is really helpful and gits fast gives the whole solution.
I write my solution after reading this hint, got AC, but it takes 1.340s which is much longer than most of others. So i think it should be solved in a o(n^3) algo. Someone has said it can be done with DP,but i can' t catch that.

mindboggler
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104 - Please explain the test case?

Post by mindboggler » Mon Jul 17, 2006 4:19 pm

4
3.1 0.0023 0.35
0.21 0.00353 8.13
200 180.559 10.339
2.11 0.089 0.06111

We are require to find the lexicographically smallest sequence giving profit greater than 1.0 right?
Then Shouldnt the answer sequence be
1 2 3 1 ?
When we start with 1 unit of currency 1, by above sequence we end up with 2.1886 unit of currency 1. That is a profit of over 100%

But the answer is 1 2 4 1

Plz Explain. Thank you

mf
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Post by mf » Mon Jul 17, 2006 4:38 pm

We are require to find the lexicographically smallest sequence
No, any minimum-length sequence will do. (So, there can be multiple correct answers.)
giving profit greater than 1.0 right?
greater than 1%.

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gradientcurl
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hi and a few qns

Post by gradientcurl » Wed Jul 19, 2006 11:12 am

f-w is already an application of DP...

btw tks abt the crash course abt f-w. but then isnt f-w only suitable for graphs with no negative weights? It is possible for a profiting sequence to exist (in some cases) if a non-simple path is taken (ie if a currency is tranverse twice). for example a->b->a->b->a (if a->b->a yields a net profit but it isnt sufficent). that would be the equivelant of a negative weight cycle if its modelled into a graph. bt f-w doesnt deal with paths which r not simple.

there's such a way of doing this using matrix multiplication, but its complexity is O(v^3lg(v)).

would appreciate if somebody can explain... maybe f-w can somehow be modified?

Darko
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Location: Calgary, Canada

Post by Darko » Wed Jul 19, 2006 4:45 pm

You can use it to detect negative cycles.
This problem is explained in Sedgewick (scroll down to "Arbitrage"):
http://www.awprofessional.com/articles/ ... Num=8&rl=1

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gradientcurl
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tip for arbitrage-> dont use logs

Post by gradientcurl » Thu Jul 20, 2006 9:40 am

ive noticed quite a number of pple used logrithms. although logs can convert multiplication into sums, thus can be used as weights of graphs, they can affect precision as one converts the weights into logs and back again. try to find ways to use multiplication to find the optimal shortest path.

abhishake
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104 - Arbitrage

Post by abhishake » Fri Sep 22, 2006 6:38 pm

Hi. I was solving 104 using brute force. I know it is not the popular method for the problem but since ive begun i want to get it done with. I seem to have a problem understanding the expected output

i/p:
4
3.1 0.0023 0.35
0.21 0.00353 8.13
200 180.559 10.339
2.11 0.089 0.06111

expected o/p:
1 2 4 1

my o/p:
1 2 3 1

The problem states that "If there is more than one sequence that results in a profit of more than 1 percent you must print a sequence of minimal length, i.e., one of the sequences that uses the fewest exchanges of currencies to yield a profit." .... now since 1 2 3 1 is also one of the minimal length sequences is it a valid o/p ?
I am further confused as 1 2 4 1 is not even the most profitable minimal length sequence (i found it to be 4 3 2 4).

Am i missing out on something or understood the q wrong ?
or will my output also be accepted ?

Thx in advance.

abhishake
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Post by abhishake » Fri Sep 22, 2006 10:16 pm

found ans in another thread .

No, any minimum-length sequence will do. (So, there can be multiple correct answers.)

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